Multiple collisions

Suppose that the two particles have velocities $u_j$ and $v_j$ toward the wall before the $j$ th collision (which takes place because $u_j > v_j$ ). Then the velocities after the $j$ th collision are determined by conservation of momentum and Newton's Law of Restitution, with $Mu_j + mv_j \; = \; Mu_{j+1} - mv_{j+1} \hspace{2cm} u_j - v_j \; = \; -u_{j+1} - v_{j+1}$ so that $u_{j+1} \; = \; \tfrac{1}{M+m}\big[(M-m)u_j + 2mv_j\big] \hspace{2cm} v_{j+1} \; = \; \tfrac{1}{M+m}\big[(M-m)v_j - 2Mu_j\big]$

If we define $x_j = \sqrt{M}u_j$ , $y_j = \sqrt{m}v_j$ and consider the acute angle $\alpha$ where $\tan\alpha = \sqrt{\tfrac{m}{M}}$ , we obtain $\left(\begin{array}{c} x_{j+1} \\ y_{j+1}\end{array}\right) \; = \; \left( \begin{array}{cc} \cos2\alpha & \sin2\alpha \\ -\sin2\alpha & \cos2\alpha \end{array}\right) \left(\begin{array}{c} x_j \\ y_j \end{array}\right)$ so that $\left(\begin{array}{c} x_j \\ y_j \end{array}\right) \; = \; \left( \begin{array}{cc} \cos2\alpha & \sin2\alpha \\ -\sin2\alpha & \cos2\alpha \end{array}\right)^{j-1} \left(\begin{array}{c} \sqrt{M}V \\ 0 \end{array}\right) \; = \; \left( \begin{array}{c} \sqrt{M}V\cos2(j-1)\alpha \\ - \sqrt{M}V\sin2(j-1)\alpha \end{array} \right)$ so long as collisions take place. For $j > 1$ , the $j$ th collision will occur provided that $v_j < 0$ (so the mass $m$ heads towards the wall after the previous collision), and also provided that $u_j > v_j$ . Thus we require $\sin2(j-1)\alpha > 0$ and $0 \; < \; \sqrt{Mm}(u_j - v_j) \; = \; \sqrt{m}x_j - \sqrt{M}y_j \; = \; \sqrt{M+m}\sin(2j-1)\alpha$ Thus the $j$ th collision will occur if $(2j-1)\alpha < \pi$ .

Thus there will be exactly $n$ collisions when $\begin{aligned} (2n-1)\alpha & < \; \pi \; \le \; (2n+1)\alpha \\ -\tfrac12\alpha & < \; \tfrac12\pi - n\alpha \; \le \; \tfrac12\alpha \\ -\tfrac12\tan^{-1}\sqrt{\tfrac{m}{M}} & < \;\tfrac12\pi - n\tan^{-1}\sqrt{\tfrac{m}{M}} \; \le \; \tfrac12\tan^{-1}\sqrt{\tfrac{m}{M}} \\ -\tfrac12\tan^{-1}x^{-\frac12} & < \; \tfrac12\pi - n\tan^{-1}x^{-\frac12} \; \le \; \tfrac12\tan^{-1} x^{-\frac12} \end{aligned}$ so that $\lim_{x \to \infty} nx^{-\frac12} \; = \; \lim_{x \to \infty} n\tan^{-1}x^{-\frac12} \; = \; \boxed{\tfrac12\pi}$

Let $u_i$ and $v_i$ be the velocities of smaller and larger particles respectively after the $i^{th}$ collision. Conservation of momentum gives,

$u_{i+1}+xv_{i+1}=-u_i+xv_i$

i.e.

$v_{i+1}-v_i=x(u_i+u_{i+1})$

which in the continuum limit as as $x$ goes to infinity and writing $t$ for the number of collisions so far gives

$\frac{dv}{dt}=\frac{2u}{x}$

similarly conservation of energy gives

$(u_{i+1}+u_i)(u_{i+1}-u_i)=x(v_{i+1}+v_i)(v_{i+1}-v_i)$

and subbing in the rearranged momentum equation gives

$u_{i+1}-u_i=v_{i+1}+v_i$

which in the limit becomes

$\frac{du}{dt}=-2v$ .

Differentiating the first differential equation and subbing the second one in gives

$\frac{d^2v}{dt^2}=-\frac{4v}{x}$

for which the solutions are of the form

$A\sin(\frac{2}{\sqrt{x}}t)+B\cos(\frac{2}{\sqrt{x}}t)$

as u starts at zero the derivative of v starts at zero so we have $A=0$ . The process finishes only when the larger particle has more speed than the smaller which won't happen until it has almost all the kinetic energy i.e. is back to its original speed, this will happen when

$\cos(\frac{2}{\sqrt{x}}t)=-1$

i.e.

$\frac{2}{\sqrt{x}}t=\pi$

i.e.

$\frac{t}{\sqrt{x}}=\frac{\pi}{2}$ .

Let us consider the most general case, when both balls are moving. Before the collision, the velocity of the ball of mass $m$ is $v$ ; the other ball's the velocity is $V$ . (The velocity are considered positive if the ball moves towards the wall.) After the collision the velocities are $v'$ and $V'$ . Momentum and energy conservation yield

$mv+MV=mv'+MV'$

$\frac{1}{2}mv^2+\frac{1}{2}MV^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2$

This can be solved for the velocities after the collision

$v'=\frac{v(m-M)+2MV}{m+M}$

$V'=\frac{V(M-m)+2mv}{m+M}$

There is one more twist: before the next collision the small ball bonces back from the wall and therefore the initial conditions for the next collision are $V=V'$ and $v=-v'$ .

Let us introduce the new variables $x=\sqrt{m}v, y=\sqrt{M}V, x'=\sqrt{m}v', y'=\sqrt{M}V'$ . Energy conservation tells us that if we plot the the $x,y$ pairs before and after the collision the quantity $x^2+y^2$ is constant. Therefore each collision corresponds to a dot on a circle in the $x-y$ plane, centered at (0,0).
The corresponding equations in terms of the new variables are

$x'=-\frac{x(m-M)+2 \sqrt{mM}y}{m+M}$

$y'=\frac{y(M-m)+2 \sqrt{mM}x}{m+M}$

In the first equation we included the "-" sign due to the collision with the wall.

To generate the dots we programmed these formulae to an Excel spreadsheet. The Figure below shows the dots for $M/m=36$ and initial velocity $V=1m/s$

The corresponding numerical values are here:

The first collision corresponds to the point at coordinates (0,6). In the 6th collision the velocity of the heavy ball changes sign, it starts to move away from the wall. However, the light ball has sufficient velocity to catch up with the heavy ball. After the 10th collision the two balls move away from the wall, and the heavy ball has slightly larger velocity. The means there will be no more collisions. On the $x-y$ plot this corresponds to the point at (-1.01,-5.91).

The Figure suggests that collisions happen as long as the dot is on the $x<0$ half plane. This is not exactly correct, because at some values of $M/m$ the "velocity check" described in the previous paragraph will differ by 1 collision. This will not matter in the $M/m\rightarrow \infty$ limit.

The Figure also suggests that the "angle" between the subsequent dots is the same all over the place. That is, in fact, true.

Let us connect two subsequent dots to the origin, and calculate the angle between the two lines. For the angle between the red line and the vertical axis we have $\sin \alpha=-x/r$ and $\cos \alpha=y/r$ . For the angle between the black line and the vertical axis we have $\sin \alpha'=-x'/r$ and $\cos \alpha'=y'/r$ . Here we needed a negative sign for the $x$ coordinates to compensate for the negative values, and $r=\sqrt{x^2+y^2}=\sqrt{x'^2+y'^2}$ . For the angle $\theta$ between the two lines we have

$\sin \theta= \sin (\alpha'-\alpha)= \frac{\sin \alpha' \cos \alpha- \cos \alpha' \sin \alpha'}{r^2}=\frac{-x'y+y'x}{x^2+y^2}$

When we substitute $x'$ and $y'$ from the elastic collision formula we discover that the angle is always the same, independent of $x$ and $y$ :

$\sin \theta=\frac{[xy(m-M)+2 \sqrt{mM} y^2]+[xy(M-m)+2 \sqrt{mM} x^2]}{(m+M)(x^2+y^2)} =\frac{2 \sqrt{mM} (y^2+x^2)}{(m+M)(x^2+y^2)}= 2 \frac{\sqrt{mM}}{m+M}$

The magnitude of the angle is

$\theta= \sin^{-1} \frac{2\sqrt{mM}}{m+M}$

We are interested In the $x=M/m\rightarrow \infty$ limit, when $\theta \approx \sin \theta \approx 2\sqrt{\frac{m}{M}}= 2 \sqrt{1/x}$ . From the $n \theta \le \pi$ condition we get

$n= \frac{\pi}{\theta}= \sqrt{x} \frac{\pi}{2}$

Therefore the answer is $\frac{n}{\sqrt{x}} = \frac{\pi}{2}$ .

Notes:

1. A more careful consideration results in the number of collisions in the form of $n=INT( \pi/\theta +0.5)$ , where $INT()$ is the floor function. The figure below shows $n/\sqrt{x}$ for $1< x<100$ . In the $x \rightarrow \infty$ limit we get the same result, $\frac{n}{\sqrt{x}} = \frac{\pi}{2}$ .

2. A similar problem has been posted recently, with an incomplete solution, see https://brilliant.org/problems/pi-day-special/

3. A paper authored by Gregory Galperin, "Playing Pool With π (The number π from a billiard point of view)", REGULAR AND CHAOTIC DYNAMICS, V. 8, №4, 2003 also discusses a similar problem. The paper can be found here http://www.turpion.org/php/full/infoFT.phtml?journalid=rd&paperid=252 or elsewhere on the WEB. The solution I presented here is different from the solution described in the paper.

4. (Added March 20, 2018) Mark Hennings's solution uses the parameter $\alpha=\tan^{-1} \sqrt{m/M}$ , see below. I used $\theta= \sin^{-1} \frac{2\sqrt{mM}}{m+M}$ . It is easy to show, by using trigonometric identities, that $\theta=2\alpha$ .

My solution is similar to that of Mark, but I thought it might be worth posting it as an alternative and to show some details.

Use $u,v$ for the velocities of the small vs. large block. Positive direction is to the right. Let $a = 1/\sqrt x = \sqrt{m/M}$ .

First, I realized that the energy $\tfrac12mu^2 + \tfrac12Mv^2$ of the system is conserved; after dividing by $\tfrac12m$ this becomes $u^2 + (v/a)^2 = \text{constant}.$ If we take $u$ and $v/a$ to describe the system, then the trajectory is limited to a circle. Anticipating that the collision will be a linear transformation of these coordinates, this means we will be dealing with rotations in the $(u,v/a)$ -plane.

Next, each "step" in this system consists of a collision between the blocks, followed by the bounce of the small block off the left wall. The center-of-mass velocity of the system is $v_{CM} = \frac{vx+u}{x+1};$ the velocities after the collision and the bounce become $u' = -(2v_{CM} - u) = \frac{x-1}{x+1}u - \frac{2x}{x+1}v; \ \ \ \ \ v' = 2v_{CM} - v = \frac{2}{x+1}u + \frac{x-1}{x+1}v.$ In matrix notation, and translated to our coordinates $(u,v\sqrt x)$ we have $\begin{pmatrix} u' \\ v'/a \end{pmatrix} = \begin{pmatrix} \cos\phi & \sin \phi \\ -\sin\phi & \cos\phi \end{pmatrix}\begin{pmatrix} u \\ v/a \end{pmatrix},$ $\text{with}\ \ \cos \phi = \frac{x-1}{x+1} \ \ \ \ \sin\phi = \frac{2\sqrt x}{x+1}.$ Thus in the $(u,v/a)$ -plane, each "step" corresponds to a rotation over angle $\phi$ .

Initially, the system is at $(0,-V/a)$ , with direction angle $-\tfrac12\pi$ ; the collisions stop when $v \approx u$ , at $(u_f,u_f/a)$ with direction angle $\tfrac12\pi - \tan^{-1} a$ . Thus the entire process corresponds to a sequence of rotations over angle $\phi$ , adding up to a total of $\pi -\tan^{-1} a$ . The number of collisions is $n = \left\lceil \frac{\pi - \tan^{-1} a}{\phi} \right\rceil.$ Now $\cos \phi = \frac{x-1}{x+1} \approx 1 - 2/x = 1 - \tfrac12(2a)^2$ when $a \to 0, x \to \infty$ ; with the small-angle approximation of the cosine, this shows that $\phi \to 2a$ when $a \to 0$ . Also, $\tan^{-1} a \to a$ in this approximation. Thus $\frac{n}{\sqrt x} \to \frac{\pi - a}{2a\sqrt x} = \tfrac12(\pi - a) \to \boxed{\pi/2 \approx 1.57}.$

This is more of a geometrical solution, speeds of the block of mass m and M will be written respectively as v and V, speeds will be assumed positive if they are oriented towards the wall.

Conservation of energy (at any time, assuming the total energy is 1/2) gives:

$v^2+xV^2=1$

which is a unit circle in the ( $v,\sqrt{x}V$ )-space

Conservation of momentum for one collision gives:

$v+xV=constant$

which is a straight line of slope $-1/\sqrt{x}$ in the ( $v,\sqrt{x}V$ )-space. As a straight line crosses a circle at most twice, at every collision between the blocks, the system switches states between the two possible states (the crossings between line and circle) and it switches from up to down as the speed of the object of mass M may only decrease. When the small object collides with the wall, its speed changes sign and that corresponds to the state mirrored across the $\sqrt{x}V$ -axis. Then once again, the next state is found by following a straight line of slope $-1/\sqrt{x}$ etc. until we can't go further. If we would concatenate all these state-graphs one would obtain something as shown on the picture.

The green line reaches the bottom of the figure at a distance $2\sqrt{x}$ on the horizontal axis. The question now becomes: how many circles are crossed when $x$ goes to infinity.

In that case, the green line can be considered almost horizontal and, as it is a straight line, it spends the same amount of time at every height, therefore, the number of circles crossed at a certain height ( $N_i$ ) is inversely proportional to the cross-width ( $w_i$ ) of the circle at that height. Hence:

$N_i w_i = c = constant$

Eventually, by taking "classes of heights" of size $h$ , we have $\frac{2}{h}$ classes and thus, the total horizontal length can be given as:

$\sum N_i w_i = \frac{2c}{h} = 2 \sqrt{x}$

Using this, we can compute $c$ and thus $N_i$ , leading us straight to $n$ as:

$n=\sum N_i = \sum \frac{h \sqrt{x}}{w_i}$

When $h$ goes to $0$ this is equivalent to:

$\frac{n}{\sqrt{x}}=\int_{-1}^1 \frac{1}{w(x)} dx = \int_{-1}^1 \frac{1}{2\sqrt{1-x^2}} dx=\frac{1}{2} [arcsin(x)]_{-1}^1$

And we find $\frac{n}{\sqrt{x}}=\frac{\pi}{2}$